Recent content by highc

This is a grade 12 advanced correspondence course, it's ultra condensed, poorly presented and getting assistance from a teacher is an obstical course of sorts. This is why I've nominated the fine folks here at physicsforums my honorary teachers. Thanks for the help on this one, I'm actually...

The vertical component of the force applied by the parent. Which will be 348.85 N (mg - F(n)). The horizontal component of force applied by the parent will equal F(f) since there is no acceleration, which is 36.7 N. With these values I can determine the net force applied by the parent to be...

One more try here... Since, F(a) = -F(f) Can I use F(f) = (W = Fd) 2200 J = F(60 m) F(f) = 36.7 N (or -36.7 N) Since, F(f) = u(k)F(n) 36.7 N = (0.26)F(n) F(n) = 141.15 N Since acceleration = 0, then F(net) = 0. Would F(a) = F(n)? Am I still lost here? This question seems as though it should...

I see what you're saying OlderDan, thank you for pointing out that error. I'm thinking F(n) should be (mg - F(a) sin theta). Do I not need the value of theta in order to do anything with this? Every way I think about this question brings me deep into the seedy underworld of theta. I can't seem...

"Work" problem I'm sure this is a straight forward problem, but I think that I may be taking the wrong approach. Any guidence would be appreciated. Problem: In many neighbourhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined...

I'm a little unsure with my some of my answers here, mainly because it seems too simple. The Questions ask: A race car driver is driving her car at the record breaking speed of 225 km/h. The first turn on the course is banked at 15 degrees, and the car's mass is 1450 kg. a) Calculate the...

Thanks alot, I wasted way too much time on this one.

I'm working through a problem that was posted here a while ago (https://www.physicsforums.com/showthread.php?t=115170). maltesers posted the equation (Gm)/(r-S)^2 = (GM)/r^2, so far this makes perfect sense, he then goes on to solve for S with S = r - (mr^2)/M, and it is here that he loses me...

Actually, after looking this over once more, I'm thinking: F(n) = F(g) cos theta + F(a) Since, F(s) = u(s)F(n), and I've determined F(s) to be 152 N, then 152 N = u(s) (F(g) cos theta + F(a)) = 0.78(152 N + F(a)) = 119 N + 0.78(F(a)) 152 N - 119 N = 0.78(F(a)) 33 N =...

This one's killing me: A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are u(s) = 0.78 and u(k) = 0.65. Determine the magnitude of the smallest force that can be applied onto the top of the...

Ok, I'm using Schaum's Outline of College Physics to supplement my ultra condensed correspondance physics course. As I work through the "Supplementary Problems" I've come across this one which leaves me puzzled. Typically a bullet leaves a standard 45 caliber pistol (5.0 inch barrel) at a speed...

The force that the lower box applies to the other is the normal force equal and opposite to the force of gravity. Would that be 9.8 N*kg?

A small box resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box. If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the...

Anybody?

Thank for the help. If anybody is ambitious enough to work this one out, can you confirm if the correct answer is: 0.41 m/s^2 [17 Degrees N of E] (I arrived at this using for B, v(X) = 7.8cos25, v(y) = 7.8sin 25. For A, v(x) = 4.4cos31, v(y) = 4.4sin31. Then, a(x) = (7.1-3.8)/8.5 = 0.39, and...